Sequences and Series jee mains questions | Sequences and Series Previous Year Question

Q 31 :
Let $a_{n}$ be the $n^{th}$ term of an A.P. If $S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700, a_{6} = 7 and S_{7} = 7$ , then $a_{n}$ is equal to [2025]

65
70
56
64

1

2

3

4

(4)

We have, $S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700, a_{6} = 7 and S_{7} = 7$

Now, $a_{6} = 7 \Rightarrow a + 5 d = 7$ ... (i)

$S_{7} = 7 \Rightarrow \frac{7}{2} (2 a + 6 d) = 7$

$\Rightarrow a + 3 d = 1$ ... (ii)

On solving (i) and (ii), we get

d = 3 and a = – 8

So, $700 = \frac{n}{2} [- 16 + (n - 1) 3]$ $[∵ S_{n} = \frac{n}{2} [2 a + (n - 1) d]]$

$\Rightarrow 1400 = - 16 n + 3 n^{2} - 3 n$

$\Rightarrow 3 n^{2} - 19 n - 1400 = 0$

$\Rightarrow (3 n + 56) (n - 25) = 0 \Rightarrow n = 25$

$∴ a_{25} = a + 24 d = - 8 + 24 \times 3 = 64$ .

Q 32 :
Suppose that the number of terms in an A.P. is 2k, $k \in N$ . If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : [2025]

6
4
5
8

1

2

3

4

(3)

Let the A.P. be $a_{1}, a_{2}, a_{3} . . . a_{2 k}$

Given, $\sum_{r = 1}^{k} a_{2 r - 1} = 40, \sum_{r = 1}^{k} a_{2 r} = 55 and a_{2 k} - a_{1} = 27$

$\Rightarrow \frac{k}{2} [2 a_{1} + (k - 1) 2 d] = 40$ ... (i)

and $\frac{k}{2} [2 a_{2} + (k - 1) 2 d] = 55$ ... (ii)

and $a_{1} + (2 k - 1) d - a_{1} = 27$ ... (iii)

After solving, we get

From (i), $a_{1} = \frac{40}{k} - (k - 1) d$

From (ii), $a_{2} = \frac{55}{k} - (k - 1) d$

and from (iii), $d = \frac{27}{2 k - 1}$

(ii) –(i)

$\Rightarrow d = \frac{15}{k} \Rightarrow \frac{27}{2 k - 1} = \frac{15}{k} \Rightarrow 9 k = 10 k - 5$

$∴$ k = 5.

Q 33 :
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to [2025]

–120
–1200
–1020
–1080

1

2

3

4

(4)

Let first term be 'a' and common difference be 'd'.

Also, $S_{4} = \frac{1}{5} (S_{8} - S_{4})$

$\Rightarrow 5 S_{4} = S_{8} - S_{4} \Rightarrow S_{8} = 6 S_{4}$

$\Rightarrow \frac{8}{2} [2 \times 3 + 7 \times d] = 6 \times \frac{4}{2} [2 \times 3 + 3 \times d]$ [ $∵ a = 3$ ]

$\Rightarrow 6 + 7 d = 18 + 9 d \Rightarrow 2 d = - 12 \Rightarrow d = - 6$

$∴ S_{20} = \frac{20}{2} [2 \times 3 + (19) \times - 6]$

= 10[– 108] = – 1080.

Q 34 :
In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57$ , then $S_{30} - S_{10}$ is equal to : [2025]

505
525
510
515

1

2

3

4

(4)

Given $S_{40} = 1030$ and $S_{12} = 57$

Let a be the first term and d be the common difference. We have,

$S_{40} = \frac{40}{2} [2 a + 39 d] = 1030; S_{12} = \frac{12}{2} [2 a + 11 d] = 57$

On simplifying, we get

$2 a + 39 d = \frac{103}{2}$ ... (i)

$2 a + 11 d = \frac{19}{2}$ ... (ii)

Subtracting (ii) from (i), we get

$28 d = 42 \Rightarrow d = \frac{3}{2}$

Using equation (ii),

$2 a + 11 \times \frac{3}{2} = \frac{19}{2} \Rightarrow a = \frac{- 7}{2}$

Now, we have

$S_{30} - S_{10} = \frac{30}{2} [2 a + 29 d] - \frac{10}{2} [2 a + 9 d]$

. $= 20 a + 390 d = 20 \times (- \frac{7}{2}) + 390 \times \frac{3}{2} = 515$ .

Q 35 :
Let $T_{r}$ be the $r^{th}$ term of an A.P. If for some m, $T_{m} = \frac{1}{25}, T_{25} = \frac{1}{20}$ and $20 \sum_{r = 1}^{25} T_{r} = 13$ , then $5 m \sum_{r = m}^{2 m} T_{r}$ is equal to [2025]

112
126
142
98

1

2

3

4

(2)

We have, $T_{m} = \frac{1}{25}, T_{25} = \frac{1}{20}$ and $20 \sum_{r = 1}^{25} T_{r} = 13$

$\Rightarrow T_{m} = a + (m - 1) d = \frac{1}{25}$ ... (i)

and $T_{25} = a + 24 d = \frac{1}{20}$ ... (ii)

Also, $20 \sum_{r = 1}^{25} T_{r} = 13$

$\Rightarrow 20 [\frac{25}{2} (2 a + 24 d)] = 13$

$\Rightarrow 20 [\frac{25}{2} (a + \frac{1}{20})] = 13$ (Using eqn. (ii))

$\Rightarrow a = \frac{1}{500}, d = \frac{1}{500}$

Using (i), we get m = 20

Now, $5 m \sum_{r = m}^{2 m} T_{r} = 5 \times 20 \sum_{r = 20}^{40} T_{r} = 126$ .

Q 36 :
Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its $11^{th}$ term is : [2025]

84
90
108
122

1

2

3

4

(2)

We have, $a_{1} + a_{2} + a_{3} = 54$

$\Rightarrow 3 (a + d) = 54 \Rightarrow a + d = 18$

$S_{20} = 10 [2 a + 19 d] = 10 [36 + 17 d]$

As $1600 < S_{20} < 1800$

$\Rightarrow 1600 < 10 (36 + 17 d) < 1800$

$\Rightarrow 160 < 36 + 17 d < 180$

$\Rightarrow 124 < 17 d < 144$

$\Rightarrow 7 \frac{5}{17} < d < 8 \frac{8}{17}$

Since, A.P. is of positive integers.

Common difference will be a natural number

$∴ d = 8 \Rightarrow a = 10$

$∴ a_{11} = 10 + 10 \times 8 = 90$ .

Q 37 :
The roots of the quadratic equation $3 x^{2} - p x + q = 0$ are $10^{th}$ and $11^{th}$ terms of an arithmetic progression with common difference $\frac{3}{2}$ . If the sum of the first 11 terms of this arithmetic progression is 88, then q – 2p is equal to __________. [2025]

0%

(474)

We have, $S_{11} = \frac{11}{2} (2 a + 10 d) = 88$

$\Rightarrow a + 5 d = 8$

$\Rightarrow a = 8 - 5 \times \frac{3}{2} = \frac{1}{2}$ $[∵ d = \frac{3}{2}]$

Let $α, β$ are the roots of the given quadratic equation.

$∴ α = T_{10} = a + 9 d = \frac{1}{2} + 9 \times \frac{3}{2} = 14$

and $β = T_{11} = a + 10 d = \frac{1}{2} + 10 \times \frac{3}{2} = \frac{31}{2}$

Sum of roots = $\frac{p}{3} = T_{10} + T_{11} = 14 + \frac{31}{2} = \frac{59}{2} \Rightarrow p = \frac{177}{2}$

and product of roots = $\frac{q}{3} = T_{10} \times T_{11} = 7 \times 31 = 217$

$\Rightarrow q = 651$ .

Now, q – 2p = 651 – 177 = 474.

Q 38 :
The interior angles of a polygon with n sides, are in an A.P. with common difference $6^{o}$ . If the largest interior angle of the polygon is $219^{o}$ , then n is equal to __________. [2025]

0%

(20)

Sum of interior angles $= (n - 2) \times 180^{o}$

$\Rightarrow \frac{n}{2} (2 a + (n - 1) 6) = (n - 2) \times 180$

$\Rightarrow a n + 3 n^{2} - 3 n = (n - 2) \times 180$ ... (i)

Now, according to question a + (n – 1)6 = 219

$\Rightarrow a = 225 - 6 n$ ... (ii)

Putting value of $a$ from equation (ii) to (i), we get

$225 n - 6 n^{2} + 3 n^{2} - 3 n = 180 n - 360$

$\Rightarrow 3 n^{2} - 42 n - 360 = 0$

$\Rightarrow n^{2} - 14 n - 120 = 0$

$\Rightarrow n = 20, - 6$ (Rejected)

Q 39 :
Let $a_{1}, a_{2}, . . ., a_{2024}$ be an Arithmetic Progression such that $a_{1} + (a_{5} + a_{10} + a_{15} + . . . + a_{2020}) + a_{2024} = 2233$ . Then $a_{1} + a_{2} + a_{3} + . . . + a_{2024}$ is equal to __________. [2025]

0%

(11132)

We have,

$a_{1} + a_{5} + a_{10} + a_{15} + . . . + a_{2020} + a_{2024} = 2233$ ... (i)

Since, $a_{1}, a_{2}, . . ., a_{2024}$ be in A.P.

$∴ a_{1} + a_{2024} = a_{5} + a_{2020} = a_{10} + a_{2015} . . . . .$

{Sum of terms equidistant form ends is equal}

From (i), we get

$(a_{1} + a_{2024}) + (a_{5} + a_{2020}) + (a_{10} + a_{2015}) + . . . = 2233$

$\Rightarrow 203 (a_{1} + a_{2024}) = 2233$ [ $∵$ Total pairs = 203]

$\Rightarrow a_{1} + a_{2024} = 11$

$∴$ Sum of n terms in A.P. is

$S_{n} = \frac{n}{2} (a_{1} + a_{2024})$

$= 1012 \times 11$ [ $∵$ n = 2024]

$∴$ $S_{n} = 11132$ .

Topic Question Set

Q 31 :
Let $a_{n}$ be the $n^{th}$ term of an A.P. If $S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700, a_{6} = 7 and S_{7} = 7$ , then $a_{n}$ is equal to [2025]

Q 32 :
Suppose that the number of terms in an A.P. is 2k, $k \in N$ . If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : [2025]

Q 33 :
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to [2025]

Q 34 :
In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57$ , then $S_{30} - S_{10}$ is equal to : [2025]

Q 35 :
Let $T_{r}$ be the $r^{th}$ term of an A.P. If for some m, $T_{m} = \frac{1}{25}, T_{25} = \frac{1}{20}$ and $20 \sum_{r = 1}^{25} T_{r} = 13$ , then $5 m \sum_{r = m}^{2 m} T_{r}$ is equal to [2025]

Q 36 :
Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its $11^{th}$ term is : [2025]

Q 37 :
The roots of the quadratic equation $3 x^{2} - p x + q = 0$ are $10^{th}$ and $11^{th}$ terms of an arithmetic progression with common difference $\frac{3}{2}$ . If the sum of the first 11 terms of this arithmetic progression is 88, then q – 2p is equal to __________. [2025]

0%

Q 38 :
The interior angles of a polygon with n sides, are in an A.P. with common difference $6^{o}$ . If the largest interior angle of the polygon is $219^{o}$ , then n is equal to __________. [2025]

0%

Q 39 :
Let $a_{1}, a_{2}, . . ., a_{2024}$ be an Arithmetic Progression such that $a_{1} + (a_{5} + a_{10} + a_{15} + . . . + a_{2020}) + a_{2024} = 2233$ . Then $a_{1} + a_{2} + a_{3} + . . . + a_{2024}$ is equal to __________. [2025]

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Topic Question Set

Q 31 : Let an be the nth term of an A.P. If Sn=a1+a2+a3+...+an=700, a6=7 and S7=7, then an is equal to [2025]

Q 32 : Suppose that the number of terms in an A.P. is 2k, k∈N. If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : [2025]

Q 33 : If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to [2025]

Q 34 : In an arithmetic progression, if S40=1030 and S12=57, then S30–S10 is equal to : [2025]

Q 35 : Let Tr be the rth term of an A.P. If for some m, Tm=125, T25=120 and 20∑r=125Tr=13, then 5m∑r=m2mTr is equal to [2025]

Q 36 : Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is : [2025]

Q 37 : The roots of the quadratic equation 3x2–px+q=0 are 10th and 11th terms of an arithmetic progression with common difference 32. If the sum of the first 11 terms of this arithmetic progression is 88, then q – 2p is equal to __________. [2025]

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Q 38 : The interior angles of a polygon with n sides, are in an A.P. with common difference 6o. If the largest interior angle of the polygon is 219o, then n is equal to __________. [2025]

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Q 39 : Let a1,a2,...,a2024 be an Arithmetic Progression such that a1+(a5+a10+a15+...+a2020)+a2024=2233. Then a1+a2+a3+...+a2024 is equal to __________. [2025]

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Q 31 :
Let $a_{n}$ be the $n^{th}$ term of an A.P. If $S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700, a_{6} = 7 and S_{7} = 7$ , then $a_{n}$ is equal to [2025]

Q 32 :
Suppose that the number of terms in an A.P. is 2k, $k \in N$ . If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : [2025]

Q 33 :
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to [2025]

Q 34 :
In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57$ , then $S_{30} - S_{10}$ is equal to : [2025]

Q 35 :
Let $T_{r}$ be the $r^{th}$ term of an A.P. If for some m, $T_{m} = \frac{1}{25}, T_{25} = \frac{1}{20}$ and $20 \sum_{r = 1}^{25} T_{r} = 13$ , then $5 m \sum_{r = m}^{2 m} T_{r}$ is equal to [2025]

Q 36 :
Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its $11^{th}$ term is : [2025]

Q 37 :
The roots of the quadratic equation $3 x^{2} - p x + q = 0$ are $10^{th}$ and $11^{th}$ terms of an arithmetic progression with common difference $\frac{3}{2}$ . If the sum of the first 11 terms of this arithmetic progression is 88, then q – 2p is equal to __________. [2025]

Q 38 :
The interior angles of a polygon with n sides, are in an A.P. with common difference $6^{o}$ . If the largest interior angle of the polygon is $219^{o}$ , then n is equal to __________. [2025]

Q 39 :
Let $a_{1}, a_{2}, . . ., a_{2024}$ be an Arithmetic Progression such that $a_{1} + (a_{5} + a_{10} + a_{15} + . . . + a_{2020}) + a_{2024} = 2233$ . Then $a_{1} + a_{2} + a_{3} + . . . + a_{2024}$ is equal to __________. [2025]